OFFSET
1,1
COMMENTS
Necessarily m = 0 (mod 3) because: (a) if m = 3k+1 then 2*m^2 + 25 = 2*(3k+1)^2 + 25 = 2*(9*k^2 + 6*k + 1) + 25 = 18*k^2 + 12*k + 27 = 0 mod 3; (b) if m = 3k-1 then 2*m^2 + 25 = 2*(3k-1)^2 + 25 = 2*(9*k^2 - 6*k + 1) + 25 = 18*k^2 - 12*k + 27 = 0 mod 3; (c) while m = 3k then 2*m^2 + 25 = 2*(3k)^2 + 25 = 18*k^2 + 25 = 1 mod 3, which can be prime. - Jonathan Vos Post, Oct 06 2005
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
EXAMPLE
If m=96 then (2*m^2) + 25 = 18457 (prime).
MATHEMATICA
Select[Range[600], PrimeQ[2 #^2+25]&] (* Harvey P. Dale, Jul 29 2025 *)
PROG
(Magma) [n: n in [0..600] | IsPrime(2*n^2+25)]; // Vincenzo Librandi, Nov 13 2010
(PARI) is(n)=isprime(2*n^2+25) \\ Charles R Greathouse IV, Jun 13 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Parthasarathy Nambi, Oct 05 2005
EXTENSIONS
More terms from Jonathan Vos Post, Oct 06 2005
STATUS
approved
